If anybody can help me with this, please do so as it is very important to me! Thank you!

1. For the function f (x) = 0.2(x4 + 4x3 - 16x - 16) + 5 complete the following table. (You may use Desmos or other graphing technology to help you. Be sure to include your graph image with your submission.) X Slope of tangent Rate of change Local Max/Min (+1-) Inflection pt. -4 - 1 2 3 4 2. A science project studying catapults sent a projectile into the air with an initial velocity of 24.5 m/s. The formula for distance (s) in meters with respect to time in seconds is s(t) = -4.9t2 + 24.5t a. Find the time that this projectile would appear to have the maximum distance above ground. (Note that you can use graphing technology to help with this, but you should also be able to analyze the problem algebraically.) b. Find the slope of the tangent at that point using lim J (x+ 7) - f(x)1. For the function f(x) = x3 , find the average rate of change (slope of secants) over each of the following intervals: a. x = 2 to x = 3 b. x = 2 to x = 2.5 C. X = 2 to x = 2.1 d. x = 2 to x = 2.01 e. x = 2 to x = 2.001 2. Using that data that you collected in question 1, what do you think the slope of the tangent to f(x) = x3 at the point (2, 8) would be? Explain you answer. 3. Using an online graphing/software (Desmos/Geogebra/Excel), graph the equation f(x) = x and show the tangent at the point (2, 8). Insert this graph (both the function and the tangent line) in your assignment. (ie. show this graph) Assistance : Creating a tangent line with Desmos. 4. A science project studying catapults sent a projectile into the air with an initial velocity of 45 m/s. The formula for height (s) in meters with respect to time in seconds is s(t) = -4.9t2 + 45t. i/ Calculate the average rate of change (average velocity) of the height over the intervals listed. a. from t=1 to t=3 b. from t=2 to t=3 c. from t=2.5 to t=3 d. from t=2.9 to t=3 e. What do you think might be happening close to t=3? Why? ii/ Calculate the instantaneous rate of change (velocity) at t = 4 seconds